[Borek]

A radioactive gas A is known to react with copper(II) oxide at 500^oC.  One of the products of this reaction, B, at the same temperature, reacts with compound M (which contains 75% by mass of Al) to yield a gas C with molar mass equal to 24, and a compound containing 53% by mass Al.  Radioactive decay of one of the atoms of compound C leads to its decomposition giving some charged particles D with molar mass equal to 21, and helium atoms ³He.  Contact of gas C with water vapors gives the molecules of an alcohol E with molar mass of 38, and hydronium ions.

What compounds are represented by letters A, B, C, D, E, and M.

[Borek]

No takers?

[Corribus]

I feel bad - this problem is all lonely and neglected, so I'll take a stab at it.  I’m not sure about some of my answers, because nuclear radiolysis reactions are not really my area of expertise, but what they heck.

I will describe my thought process, then provide my answers.  Let’s see how close I got.

The most obvious place for me to start was actually molecule C.  C is radioactive like A, but here we’re given two important clues: one is the low molecular weights involved and the other is that the product of the radioactive decay is a light helium, ³He.  There really are two major possibilities here for a radioactive isotope with low molecular weight: tritium and carbon-14.  Both undergo beta decay, but only tritium gives up a light helium as well.  Therefore C and A, I conclude, both involve tritium, henceforth abbreviated as T.

Moving to A.  There are a couple of common gases that have tritium in them that might react with cupric oxide: di-tritium, tritiated ammonia, and tritiated methane.  NT₃ and T₂ both react with cupric oxide to give water, copper metal, and – in the case of ammonia – nitrogen gas.  Tritiated methane produces tritiated water and CO₂.  Not much help at the moment, since the products are somewhat similar.

Ok, I file that in the back of my mind and take a look at the reaction in the middle, the one with aluminum.  The compound with 53% by mass of aluminum is clearly aluminum oxide, Al₂O₃.  Well, it’s the first possibility I checked since it’s the most obvious.  Sadly, and most unfairly, I get no credit for this achievement because it’s not one of my unknowns.  But it does tell me that mystery compound C must contain my tritium source.  I know C must also react with water to give an alcohol (E), so therefore I conclude that C must have some carbon in it, and playing around a little with the weights to get MW = 24 gives me a convenient tritiated methane, CT₄, for compound C.  Given that this was also one of my possibilities for A, and thinking it’d be rather redundant to have the same compound for A and C, I eliminate CT₄ from my possibilities for A.

My original thought for M was a tritiated version of aluminum trihidride – very reactive and, by coincidence, conveniently having almost exactly 75% by weight of aluminum.  That coincidence threw me off for quite a while, because that meant B had to be my carbon source, which made no sense after I eliminated CT₄ from possibilities for A.  Banged my head against the wall for awhile at that one, because I was dead set on AlT₃, with the nice 75% and all, but no, M just had to have my carbon source.  No other way.  Trimethyl aluminum was my first thought here but it’s not 75% aluminum by weight - crap.  Almost emailed Borek and asserted that there was a typo, 75% couldn't be right.  LOL at my ego, thinking a typo is more likely than that I was wrong.  Well, I’ll be honest here and admit that at this point I just started to look around on Wikipedia for aluminum compounds that contain carbon.  Came up with aluminum carbide – 75% aluminum by weight and, what do you know, reacts with water to give methane and alumina.  Bingo.  Now does that make me a cheat, or just resourceful?  I'll let the chemistry gods decide.

Anyway, that means M is aluminum carbide, Al₄C₃, and that means B is tritiated water, T₂O.  Front half of the problem done.

Back half: I thought originally this would be easy until I realized I know squat about self-radiolysis reactions.  I knew C was CT₄.  Decay of one of the tritiums to the light helium knocks 3 g/mol off of CT₄’s weight, which gives D, conveniently at molar mass of 21.  D must therefore be CT₃^- or CT₃⁺.  I put aside for the moment which of these two possibilities it is and move to the last reaction:  Reaction of D with water gives an alcohol E with molar mass of 38 and some hydronium, H₃O⁺.  Well, methanol is the obvious choice for the alcohol, and tritiated methanol CT₃OH has a mass of 38, so I think this is E.  D can’t be a carbanion and produce hydronium – if D was a carbanion and it reacted with water, I surmise it would give a hydroxyl and produce CT₃H, another tritiated methane.  No go. Therefore D must be the short-lived methenium, CT₃⁺, which should nicely rip off a hydroxyl from water and produce a bunch of extra protons.

So in conclusion:

A = T₂
B = T₂O
C = CT₄
D = CT₃⁺
E = CT₃OH
M = Al₄C₃

Here are the balanced equations:

CT₄ (g) + CuO(s)  Cu (S) + T₂O (l)
6 T₂O (l) + Al₄C₃ (s)  2 Al₂O₃ (s) + 3 CT₄(g)
CT₄  CT₃⁺ + ³He + e^-
CT₃⁺ + 2H₂O  CT₃OH + H₃O⁺

Ok.. how’d I do?

[Borek]

A = T₂
B = T₂O
C = CT₄
D = CT₃⁺
E = CT₃OH
M = Al₄C₃

Hard to deny, that's the correct answer

Sadly, we have at least three Chemistry Olympiad wannabes and none of them attempted the problem 

Edit: forgot to mention, this question was already posted at CF several years ago, I found it accidentally last week while doing some database cleaning.

[Corribus]

Great, spent all that time and all I had to do was a forum search.

[Borek]

No, the old thread was soft deleted, I found it while checking what should be deleted permanently. I have undeleted it around the time I posted the link.

[Big-Daddy]

Sorry Borek!

I got as far as C is CT₄ (to use Corribus' notation), D is CT₃⁺ and E is CT₃OH.

However, without looking at Corribus' answer I was unable to work out A, B or M.

Clearly I have a long way to go in these types of problems.

[Borek]

Note: it is not just a Corribus notation, H, D and T are pretty much established as symbols for protium, deuterium and tritium.

[Corribus]

IUPAC discourages it, but it's a heck of a lot easier than writing all the subscripts and superscripts.

[Big-Daddy]

Note: it is not just a Corribus notation, H, D and T are pretty much established as symbols for protium, deuterium and tritium.

I've always written the odd isotopes out (e.g. ¹⁴C, ²H etc.)but I have seen D for deuterium many times - rarely encountered ³H i.e. T so I thought it was worth mentioning.